小区间浮动的范围函数

Question

我想使用导数的定义求函数 (x)(x - 1) 的导数。我希望我的增量为 1e-2。这样它就可以模拟极限变为零。我在 Range for Floats 上看到我可以使用用户定义的函数来创建采用浮点变量的范围函数。

def frange(x, y, jump):
    while x < y:
        yield x
        x += jump

    def drange(start, stop, step):
     r = start
     while r < stop:
        yield r
        r += step

i = frange(1e-14,1e-2,2)

for k in i:
    set  = []
    x = 1
    dvt = ((x + k ) * (x + k - 1) - x*(x - 1))/k

    set.append(dvt)
    print(set)

当我运行程序时我只得到

[0.9992007221626509]

我没有在列表中添加超过一种衍生品，这是怎么回事？

Answer 1

设置你在说什么

x += jump

这会将 x 的值设置为 2 + 1e-14，它大于 1e-2

我看了代码，好像你的意思是

myjump = pow(10, jump) #outside the loop

x *= myjump # inside the loop

这会将示例中的每个循环乘以 100 并处理 1e-14、1e-12、1e-10 ... 1e-2

或者，如果你想添加它，那么你应该说

x += myjump # inside the loop

或者你需要测试跳跃实际上是一个小到足以被处理的分数。

Answer 2

这是一个 cleaned-up 版本：

def fn(x):
    return x * (x - 1)

def numerical_diff(fn, x, delta):
    return (fn(x + delta) - fn(x)) / delta

def geometric_series(a, r, n):
    value = a
    for i in range(n):
        yield value     # a * r ** i
        value *= r

def main():
    x = 1.
    for delta in geometric_series(0.01, 0.01, 7):
        print(
            "x = {:5.3f}  f(x) = {:5.3f}  delta = {:16.14f}  f'(x) = {:16.14f}"
            .format(x, fn(x), delta, numerical_diff(fn, x, delta))
        )

if __name__ == "__main__":
    main()

产生

x = 1.000  f(x) = 0.000  delta = 0.01000000000000  f'(x) = 1.01000000000000
x = 1.000  f(x) = 0.000  delta = 0.00010000000000  f'(x) = 1.00009999999989
x = 1.000  f(x) = 0.000  delta = 0.00000100000000  f'(x) = 1.00000099991773
x = 1.000  f(x) = 0.000  delta = 0.00000001000000  f'(x) = 1.00000000392253
x = 1.000  f(x) = 0.000  delta = 0.00000000010000  f'(x) = 1.00000008284037
x = 1.000  f(x) = 0.000  delta = 0.00000000000100  f'(x) = 1.00008890058334
x = 1.000  f(x) = 0.000  delta = 0.00000000000001  f'(x) = 0.99920072216265

小区间浮动的范围函数

Range Function For Floats In Small Intervals

python

calculus

differentiation

python-3.x